Question

A bullet is fired vertically upwards with velocity $v$ from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is $\frac{1}{4}$th of its value at the surface of the planet. If the escape velocity from the planet is $v_0=v\sqrt{N}$, then the value of $N$ is
[ignore energy loss due to atmosphere]

Answer 1

At height $h$, acceleration due to gravity $g^{\prime }=\frac{g}{4}$
[given]

From the formula for acceleration due to gravity at height '$h$' above the ground,
$\frac{g}{4}=\frac{g}{{(1+\frac{h}{R})}^2}$
$\Rightarrow {(1+\frac{h}{R})}^2=4$
$\Rightarrow 1+\frac{h}{R}=2$
$\Rightarrow h=R$ ...$\left(1\right)$

On applying law of conservation of mechanical energy from $A$ to $B$,
Decrease in kinetic energy = Increase in potential energy
$K.E_B-K.E_A=P.E_B-P.E_A$
$\Rightarrow 0-\frac{1}{2}m{v}^2=m{g}^{\prime }(R+h)-mgR$
$\Rightarrow 0-\frac{1}{2}m{v}^2=m\left(\frac{g}{{(1+\frac{h}{R})}^2}\right)(R+h)-mgR$
$\Rightarrow \frac{1}{2}m{v}^2=\frac{mgh}{1+\frac{h}{R}}$
$\Rightarrow \frac{v^2}{2}=\frac{gR}{1+\frac{R}{R}}$ from $\left(1\right)$
$\Rightarrow \frac{v^2}{2}=\frac{gR}{2}$
$\Rightarrow v=\sqrt{gR}$ ...$\left(2\right)$

Now, escape velocity, $v_0=\sqrt{2gR}$
$\Rightarrow v_0=\sqrt{2}\times v$ from $\left(2\right)$
$\Rightarrow v_0=v\sqrt{2}$

From $v_0=v\sqrt{N}$ $($given$)$
$\Rightarrow N=2$

Why this question? Concept involved - $\left(1\right)$ variation of acceleration due to gravity with height or depth. $\left(2\right)$ Law of conservation of mechanical energy. Importance in $JEE$ - Asked in Advanced $2015$.